What would be a group isomorphism betweem these two?

Let $G=\{id,(1234),(1432),(13)(24),(14)(23),(12)(34),(13),(24)\}$

I have shown that $G$ is a subgroup of $S_4$.

Let $\phi:Z_2 \rightarrow Aut(Z_4)$ be a group homomorphism such that $\phi(1)(g)=g^{-1}$.

Define $Dih_4=Z_4\rtimes_\phi Z_2$.

How do I show that $G$ is isomorphic to $Dih_4$?

I have a trouble with constructing the explicit isomorphism..

• This is a bit confusing. This subgroup of order $8$ that you found in $S_4$. Do you think this is just the permutation representation of the symmetry group on the square? – Mr.Fry Jun 4 '15 at 0:31
• @Mr.Fry Every subgroup of order $8$ are mutually isomorphic in this case, so I think that doesn't matter.. And actually I have found one – Rubertos Jun 4 '15 at 0:37

2 Answers

Another way to show that $Dih_4$ is isomorphic to the group you wrote. Clearly $G$ is a non-abelian group of order $8$. I make the following claim : any non-abelian group of order $8$ with at least two elements of order $2$ is isomorphic to $Dih_4$.

Take $H$ to be such a group. First $exp(H)=2,4,8$ if $exp(H)=2$ or $8$ then $H$ is abelian. Hence $exp(H)=4$. Now take $h_0\in H$ of order $4$. Then $H_0:=<h_0>$ is of index $2$ in $H$ hence it is normal in $H$.

Now in $H_0$ we only have $1$ element of order $2$, so that there exists another $h_1$ of order $2$ outside $H_0$. Denote $H_1:=<h_1>$. Then I claim that :

$$H_0\cap H_1=\{1_H\}$$

$$H_0\triangleleft H$$

$$H=H_0H_1$$

From this it follows that $H$ cannot but be a semi direct product :

$$H=H_0\rtimes_{\phi}H_1$$

where $\phi:H_1\rightarrow Aut(H_0)$ now $Aut(H_0)=\{\pm Id\}$ and $H_1$ is the group of order $2$ hence we have two possibilities $\phi(1)=Id$ or $\phi(1)=-Id$ in the first possibility we have that $H$ is isomorphic to $\frac{\mathbb{Z}}{4\mathbb{Z}}\times \frac{\mathbb{Z}}{2\mathbb{Z}}$ hence abelian (impossible) so we are in the second case and the second case leads to $Dih_4$ directly.

Set $f(id)=(0,0), f(1234)=(1,0), f((13)(24))=(2,0), f(1432)=(3,0), f((14)(23))=(0,1), f((12)(34))=(2,1), f(13)=(3,1), f(24)=(1,1)$